Let's break down the problem and solve it step by step. --- ## **Step 1: Extract Given Information** - **Incidence rate (control group):** 150 cases per 100,000 women/year - **Relative risk reduction (expected):** 20% reduction (so risk ratio = 0.80) - **Alpha (significance level):** 0.05 (also need to calculate for 0.10) - **Power:** 85% (i.e., beta = 0.15), also need to calculate for 80% - **Ratio (case:control):** 1:1 (equal numbers in each group) --- ## **Step 2: Calculate Event Rates** - **p1 (control group incidence):** 150/100,000 = 0.0015 per year - **p2 (treatment group incidence):** 0.8 × 0.0015 = 0.0012 per year - **Expected incidence in treatment group:** 120/100,000 --- ## **Step 3: Sample Size Formula for Comparing Two Proportions** The formula for each group (when p1 and p2 are small) is: \[ n = \frac{(Z_{1-\alpha/2} + Z_{1-\beta})^2 \cdot [p_1(1-p_1) + p_2(1-p_2)]}{(p_1 - p_2)^2} \] Where: - \( p_1 \) = incidence in control - \( p_2 \) = incidence in treatment - \( Z_{1-\alpha/2} \) = critical value for significance level (two-sided) - \( Z_{1-\beta} \) = critical value for power --- ## **Step 4: Find Z-values** - **For α = 0.05:** \( Z_{1-\alpha/2} \) = 1.96 - **For α = 0.10:** \( Z_{1-\alpha/2} \) = 1.645 - **For Power 0.85:** \( Z_{1-\beta} \) = 1.036 - **For Power 0.80:** \( Z_{1-\beta} \) = 0.842 --- ### **A. Original: α = 0.05, Power = 0.85** \[ n = \frac{(1.96 + 1.036)^2 \cdot [0.0015 \times (1-0.0015) + 0.0012 \times (1-0.0012)]}{(0.0015 - 0.0012)^2} \] Calculate numerators and denominators: - \( (1.96 + 1.036)^2 = (2.996)^2 \approx 8.976 \) - \( 0.0015 \times 0.9985 = 0.00149775 \) - \( 0.0012 \times 0.9988 = 0.00119856 \) - Sum: \( 0.00149775 + 0.00119856 = 0.00269631 \) - \( (0.0015 - 0.0012)^2 = (0.0003)^2 = 0.00000009 \) Plug in: \[ n = \frac{8.976 \times 0.00269631}{0.00000009} \] \[ n = \frac{0.024224}{0.00000009} \] \[ n = 269,155 \] **So, each group needs about 269,155 women.** --- ### **B. If α = 0.10, Power = 0.85** \( Z_{1-\alpha/2} = 1.645 \), \( Z_{1-\beta} = 1.036 \) \[ (1.645 + 1.036)^2 = (2.681)^2 = 7.188 \] \[ n = \frac{7.188 \times 0.00269631}{0.00000009} = \frac{0.019393}{0.00000009} = 215,478 \] **So, each group needs about 215,478 women.** --- ### **C. If Power = 0.80, α = 0.05** \( Z_{1-\alpha/2} = 1.96 \), \( Z_{1-\beta} = 0.842 \) \[ (1.96 + 0.842)^2 = (2.802)^2 = 7.851 \] \[ n = \frac{7.851 \times 0.00269631}{0.00000009} = \frac{0.021163}{0.00000009} = 235,144 \] **So, each group needs about 235,144 women.** --- ## **Summary Table** | α | Power | Sample size per group | |------|-------|----------------------| | 0.05 | 0.85 | 269,155 | | 0.10 | 0.85 | 215,478 | | 0.05 | 0.80 | 235,144 | --- ### **Conclusion** - **a) For α = 0.05, 85% power:** **269,155** women per group (total ≈ 538,310). - **b) For α = 0.10, 85% power:** **215,478** women per group (total ≈ 430,956). - **c) For α = 0.05, 80% power:** **235,144** women per group (total ≈ 470,288). > These large sample sizes reflect the **very low incidence rate** of breast cancer over 1 year in this age group; to detect a 20% reduction with good power, a very large sample is needed. --- **If you need the calculations for a longer follow-up (e.g., 5 years), the sample size would decrease since more events would accrue.**